Answer to Question #139774 in Calculus for Moel Tariburu

Question

Evaluate the integral ∫▒(〖tan〗^4 x+〖tan〗^6 x)/(〖tan〗^4 x-1) dx

Answer 1

$\int \frac{tan^4x+tan^6x}{tan^4x-1}dx=\int \frac{tan^4x}{tan^2x-1}dx=$

$\int( tan^2x+1+\frac{1}{tan^2x-1})dx=\int(tan^2x+1)dx+\int \frac{1}{tan^2x-1}dx=$

$=tanx+\int\frac{cos^2x}{sin^2x-cos^2x}dx=tanx+\int \frac{\frac{1+cos2x}{2}}{-cos2x}dx=$

$=tanx+\int(-\frac{1}{2cos2x}-\frac12)dx=$

$=tanx-\frac14log|sec(2x)+tan(2x)|-\frac x2+C$